4
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Find the next number in this sequence:

2, 4, 8, 13, 20, 29, 41, ?

EXTENSION: Find the number for the nth term in the sequence (in terms of n).

Hint:

Fibonacci+Triangle

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  • $\begingroup$ mathworld.wolfram.com/TribonacciNumber.html $\endgroup$
    – Duck
    Commented Sep 9, 2018 at 2:33
  • $\begingroup$ The hint has nothing to do with that. $\endgroup$ Commented Sep 9, 2018 at 2:34
  • $\begingroup$ Ok, just checking $\endgroup$
    – Duck
    Commented Sep 9, 2018 at 2:35
  • $\begingroup$ And I changed it to make it more obvious. $\endgroup$ Commented Sep 9, 2018 at 2:37
  • 1
    $\begingroup$ OEIS found this but I don't think that is what you had in mind $\endgroup$ Commented Sep 9, 2018 at 2:45

2 Answers 2

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I think the answer is

57

Solution for the extension: (Thanks @PerpetualJ for MathJax)

$$Fi. Δ.$$ $$1 + 1 = 2$$ $$1 + 3 = 4$$ $$2 + 6 = 8$$ $$3 + 10 = 13$$ $$5 + 15 = 20$$ $$8 + 21 = 29$$ $$13 + 28 = 41$$ $$21 + 36 = 57 \leftarrow (Next Term)$$

Round off to the nearest integer.
$Fi(n) = round\biggl(\frac{(-\frac{1}{\phi})^n}{\sqrt{5}}\biggr)$
$\Delta(n) = \frac{n(n + 1)}{2}$
$\int(n) = Fi(n) + \Delta(n) = round\biggl(\frac{(-\frac{1}{\phi})^n}{\sqrt{5}}\biggr) + \frac{n(n + 1)}{2}$

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  • 1
    $\begingroup$ Yes, good job!! $\endgroup$ Commented Sep 9, 2018 at 7:00
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The next number is

$34$

The formula is

$2^n - \dfrac{(n-1)(n-2)(n-3)}2$

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  • $\begingroup$ I'm thinking of a different formula and I realized this worked as well so I'll edit the question. $\endgroup$ Commented Sep 9, 2018 at 2:25

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